I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 1.9 : Exponential And Logarithm Equations
6. Find all the solutions to \(14{{\bf{e}}^{6 - x}} + {{\bf{e}}^{12x - 7}} = 0\). If there are no solutions clearly explain why.
With both exponentials in the equation this may be a little difficult to solve, so let’s do some work to reduce this down to an equation with a single exponential.
\[\begin{align*}14{{\bf{e}}^{6 - x}} & = - {{\bf{e}}^{12x - 7}}\\ \frac{{{{\bf{e}}^{12x - 7}}}}{{{{\bf{e}}^{6 - x}}}} & = - 14\\ {{\bf{e}}^{13x - 13}} & = - 14\end{align*}\]At this point we can stop. We know that exponential functions are always positive and there is no way for this to be negative and therefore there is no solution to this equation.
Note that if we hadn’t caught the exponent being negative our next step would have been to take the logarithm of both side and we also know that we can only take the logarithm of positive numbers and so again we’d see that there is no solution to this equation.